9 Qudrtic Equtions Qudrtic epression nd qudrtic eqution Pure nd dfected qudrtic equtions Solution of qudrtic eqution y * Fctoristion method * Completing the squre method * Formul method * Grphicl method Reltion etween roots nd coefficients Forming qudrtic equtions This unit fcilittes you in, defining qudrtic eqution, pure nd dfected qudrtic equtions. solving pure qudrtic equtions. solving qudrtic e qutions y fctoristion method. solving qudrtic e qutions y completing the squre method. deriving the formul to find the roots of qudrtic eqution. using formul to solve qudrtic equtions. drwing grphs for qudrtic epressions. solving qudrtic equtions grphiclly. estlising reltion etween roots nd coefficients of qudrtic eqution. frming qudrtic equtions. Brhmgupt (A.D 98-66, Indi) Solving of qudrtic equtions in generl form is often credited to ncient Indin mthemticins. Brhmgupt gve n epl icit formul to solve qudrtic eqution of the form + c. Lter Sridhrchry (A.D. 0) derived formul, now known s the qudrtic formul, for solving qudrtic eqution y the method of completing the squre. finding the discriminnt nd interpret the nture of roots of qudrtic eqution. An eqution mens nothing to me, unless it epresses thought of God. - Srinivs Rmnujn
9 UNIT-9 We re fmilir with plying numer gmes. Let us consider two such emples. Emple Emple * Tke non-zero whole numer. * Tke non-zero whole numer * Add 7 to it. * Add 7 to it. * Equte it to. * Multiply the sum y the sme whole numer. * Wht is the numer? * Equte it to. * Wht is the numer? How to find the numers? Let the non-zero whole numer e '' If we follow the steps, we get in emple, + 7 nd + 7 in emple, + 7 nd ( + 7)...(i)...(ii) Tke eqution (i), + 7. Here, is the vrile, whose degree is. It is liner eqution. It hs only one root. i.e., + 7, 7, Tke eqution (ii), ( + 7), + 7 Here, is the vrile, whose degree is. It is qudrtic eqution. How to solve it? How mny roots does qudrtic eqution hve? It is very essentil to lern this, ecuse qudrtic equtions hve wide pplictions in other rnches of mthemtics, in other sujects nd lso in rel life situtions. For instnce, suppose n old ge home trust decides to uild pryer hll hving floor re of 00 sq.m., with its length one meter more thn twice its redth. Wht should e the length nd redth of the hll? Let, the redth e m. Then, its length will e ( + )m Its re ( + )sq.m ( + ) 00 This informtion cn e digrmmticlly represented s follows. We hve, Are ( + ) ( + )m So, + 00. This is qudrtic eqution. ( Given) 00 m m ( + ) m Below re given some more illustrtions in verl sttement form which when converted into eqution form result in qudrtic eqution form. Study the sttements. Try to epress ech sttement in eqution form.. An epress trin tkes one hour less thn the pssenger trin to trvel km etween Bnglore nd Mysore. If the verge speed of the epress trin is km/hr more thn tht of the pssenger trin, wht is the verge speed of the two trins?
Qudrtic Equtions 9. A cottge industry produces certin numer of wooden toys in dy. The cost of production of ech toy (in rupees) ws found to e minus the numer of toys produced in dy. On prticulr dy, the totl cost of production ws ` 70. Wht is the numer of toys produced on tht dy?. A motor ot whose speed is 8 km/hr in still wter tkes hour more to go km upstrem thn to return downstrem to the sme spot. Wht is the speed of the strem?. Two wter tps together cn fill tnk in 9 8 hours. The tp of lrger dimeter tkes 0 hours less thn the smller one to fill the tnk seprtely. Find the time in which ech tp cn seprtely fill the tnk. How to solve these prolems? Know this! It is elieved tht Bylonins were the first to solve qudrtic equtions. Greek mthemticin Euclid developed geometricl pproch for finding lengths, which re nothing ut solutions of qudrtic equtions. Solving of qudrtic equtions, in generl form, is often credited to ncient Indin mthemticins like Brhmgupt (A.D. 98-66) nd Sridhrchry (A.D. 0). An Ar mthemticin Al-khwrizni (out A.D. 800) lso studied qudrtic equtions of different types. Arhm r Hiyy H-Nsi, in his ook "Lier emrdorum" pulished in Europe in A.D. gve complete solutions of different qudrtic equtions. In this unit, let us study qudrtic equtions, vrious methods of finding their roots nd lso pplictions of qudrtic equtions. Qudrtic eqution Recll tht we hve studied out qudrtic polynomils in unit 8. A polynomil of the form + + c, where 0 is qudrtic polynomil or epression in the vrile of degree. If qudrtic epression + + c is equted to zero, it ecomes qudrtic eqution. Below re given some verl sttements, when converted to qudrtic epression form nd further equted to zero ecome qudrtic equtions. Study the emples.
96 UNIT-9 Sl. Verl sttement Qudrtic Qudrtic No. epression eqution. The sum of numer nd five + + 0 times its squres.. A wire is ent to form the legs ( of right ngled tringle. ) ( ) 0 If one of them is cm more thn the other, wht will e ( ) its re? 0. The runs scored y cricket 6 6 0 tem re 6 less thn the difference of the runs scored y the first tsmn nd the squre of runs scored y him So it cn e stted tht, if p() is qudrtic polynomil, then p() 0 is qudrtic eqution. In fct ny eqution of the form p() 0, where p() is polynomil of degree is qudrtic eqution whose stndrd form is + + c 0, 0. A qudrtic eqution in the vrile is n eqution of the form + + c 0 where,, c, re rel numers nd 0. Chrcteristics of qudrtic eqution re, Know this! it is n eqution in one vrile. it is n eqution whose single vrile is of degree. the stndrd form of qudrtic eqution is + + c 0. Here, is the coefficient of, is the coefficient of, c is the constnt term,,, c re rel numers nd 0. The word qudrtic is derived from the Ltin word "qudrtum" which mens "A squre figure". the terms re written in descending order of the power of the vrile. In qudrtic eqution why is 0? Wht hppens to the qudrtic eqution, if 0? Discuss in the clss. We hve discussed tht in qudrtic eqution, 0. Wht hppens to stndrd form of qudrtic eqution when or c or oth nd c re equl to zero?
Qudrtic Equtions 97 Oserve the tle given elow. Sl. Vlue of vlue of c Result. 0 c 0 + c 0. 0 c 0 + 0 0 c 0 0. 0 c 0 + + c 0 Oserve tht, in ll the ove cses the eqution remins s qudrtic eqution. ILLUSTRATIVE EXAMPLES Emple : Check whether the following re qudrtic equtions: (i) + + 0 (ii) 6 + (iii) ( 8) 0 0 (iv) ( + ) + 8 ( + ) ( ) Sol. (i) + + 0 Arrnge the terms in descending order of their powers. + + 0 It is in the stndrd form + +c 0. The given eqution is qudrtic eqution. (ii) 6 + (iii) By rerrnging the terms, we get 6 + 0 The highest degree of the vrile is. The given eqution is not qudrtic eqution. ( 8) 0 0 By simplifying we get 8 0 0 + + 0 0 6 + 0 0. It is of the form + + c 0. it is qudrtic eqution. (iv) ( + ) + 8 ( + )( ) By simplifying we get + + 8 It is not of the form + + c 0. The vrile is only in the first degree. it is not qudrtic eqution. + + 8 + 0, + 0
98 UNIT-9 EXERCISE 9.. Check whether the following re qudrtic equtions: (i) 0 (ii) 8 (iii) 0 (iv) 0 0 9 (v) 0 (vi) 6 (vii) 0 (viii) (i) 0 + 7 0 () y 0. Simplify the following equtions nd check whether they re qudrtic equtions. (i) ( + 6) 0 (ii) ( - )( ) 0 (iii) ( + 9)( 9) 0 (iv) ( + )( 7) (v) + ( )( 9) 0 (vi) ( + ) ( ) (vii) ( )( ) ( + )( ) (viii) + + ( ) (i) ( + ) ( ) () + ( ). Represent the following in the form of qudrtic equtions. (i) The product of two consecutive integers is 06. (ii) The length of rectngulr prk (in metres) is one more thn twice its redth nd its re is 8m. (iii) A trin trvels distnce of 80 km t uniform speed. If the speed hd een 8 km/hr less, then it would hve tken hours more to cover the sme distnce. Oserve the emples given in the tle. A 7 + 0 0 8 0 7y y + 0 0 y 0 We cn oserve, tht, B p 0 8p All the equtions in A nd B re qudrtic equtions. All the qudrtic equtions in column A hve the vrile in second degree only. They re clled pure qudrtic equtions. All the qudrtic equtions in column B hve the vrile in oth second degree nd first degree. They re clled dfected qudrtic equtions. Solution of qudrtic equtions: Consider the pure qudrtic eqution 0. Let us tke rel vlues for '' nd sustitute in the eqution. Let Let LHS LHS 0 LHS RHS LHS RHS Co mpre the equtions in columns A nd B.
Qudrtic Equtions 99 Let Let LHS ( ) LHS ( ) 0 LHS RHS LHS RHS Try this for different vlues of. We oserve tht LHS RHS only for two vlues of, i.e., nd. We sy, + nd stisfy the eqution 0. + nd re clled roots of the qudrtic eqution + nd re lso clled zeroes of the polynomil. The roots of the qudrtic eqution stisfy the eqution, which mens LHS RHS. The ove discussion holds good for dfected qudrtic equtions lso. Consider the dfected qudrtic eqution 0 0 Let Let LHS 0 () 0 LHS 0 ( ) ( ) 0 8 LHS RHS LHS RHS Let Let LHS 0 () 0 LHS 0 ( ) ( ) 0 0 0 LHS RHS LHS RHS We oserve tht the qudrtic eqution 0 0 is stisfied only for the vlues nd. nd re the roots of the qudrtic eqution 0 0 nd re the zeroes of the qudrtic polynomil 0. In generl, rel numer 'k' is clled root of the qudrtic eqution + + c 0, 0 if k + k + c 0. We lso sy tht k is solution of the qudrtic eqution, or tht k stisfies the qudrtic eqution. Note tht the zeroes of the qudrtic polynomil + + c nd the roots of the qudrtic eqution + + c 0 re the sme. We know tht qudrtic polynomil hs t most two zeroes. Any qudrtic eqution hs t most two roots. Solving qudrtic eqution mens, finding the roots of qudrtic eqution. The roots cn e verified y sustituting the vlues in the qudrtic eqution nd checking whether they stisfy the eqution. The roots lso form the solution set of the qudrtic eqution.
00 UNIT-9 Solution of pure qudrtic eqution. Solve 0 Sol. 0, + or. Solve t Sol. t Rewrite the eqution in stndrd form t +, t t +, t + t + or t. If V r h, then solve for 'r' nd find the vlue of 'r' when V 76 nd h. Sol. Given V r h Recll: Positive rel numers will hve two roots. One is +ve & the other root is ve & 0 0 r h V, r V h V r h If V 76 nd h, we get r 8 7 76 (y sustituting the vlues) r 76 7 76 r + or r. Shwet owns 6 m lnd in squre shpe. She wnts to fence the lnd with red wire. Clculte the length of the wire required for rounds. Sol. Let length of the squre e m., Are of the squre 6, 6 + or Since the length cnnot e negtive, we tke +. length of one side of the squre +m Perimeter of squre m wire required for four rounds 6 m required length of the wire 6 00 m EXERCISE 9.. Clssify the following equtions into pure nd dfected qudrtic equtions. (i) 00 (ii) + 6 6 (iii) p(p ) (iv) + (v)( + 9)( 9) 0 (vi) 7 (vii) 0 (viii) 7 (i) () 8 (i) ( ) 8 (ii) 8 9 ( )
Qudrtic Equtions 0. Solve the qudrtic equtions: (i) 96 0 (ii) 6 (iii) + 0 (iv) 7 6 7 (v) ( + 8) (vi) 7 (vii) + 0 (viii) 7. 7.. In ech of the following, determine whether the given vlues of '' is solution of the qudrtic eqution or not. (i) + + 0 ;, (ii) 7 0 ; (iii) m 6m + 0 ; m (iv) y y 0 ; y (v) (vii) ; nd (vi) (k + 8)(k + ) 0 ; k nd k ; nd (viii) 6 0 ; nd. (i) If A r, solve for r, nd find the vlue of 'r' if A 77 nd 7. (ii) If r l + d solve for d, nd find the vlue of 'd' if r nd l. (iii) If c + solve for nd find the vlue of, if 8 nd c 7. (iv) If A solve for nd find the vlue of, if A 6. (v) If k mv solve for 'v' nd find the vlue of v, if k 00 nd m. (vi) If v u + s solve for 'v' nd find the vlue of 'v', if u 0,, s 00. Solution of dfected qudrtic equtions We know tht the generl form of n dfected qudrtic eqution is + + c 0, 0. This eqution cn lso occur in different forms such s + 0, + c 0 nd 0. 0 nd + c 0 re pure qudrtic equtions nd we hve lernt in the previous section the method of solving them. How to solve n dfected qudrtic eqution? There re severl methods of solving the dfected qudrtic equtions depending on their forms, ie., + + c 0 or + 0. These methods pply for pure qudrtic equtions lso. Let us lern them. (A) Solution of qudrtic eqution y fctoristion method. We hve lernt to fctorise qudrtic polynomils y splitting their middle terms. We shll use this knowledge for finding the roots of qudrtic eqution.
0 UNIT-9 Fctoristion method is used when the qudrtic eqution cn e fctorised into two liner fctors. After fctoristion, the qudrtic eqution is epressed s the product of its two liner fctors nd this is equted to zero. Tht is, + + c 0 nd ( m) ( n) 0 Then, we pply zero product rule nd equte ech fctor to zero nd solve for the unknown. Zero product rule: i.e., m 0 m Let nd e ny two rel n 0 n numers or fctors. If 0 then, either 0 or 0 oth nd re equl to zero So, m nd n re the roots of the qudrtic eqution + + c 0. Consider the qudrtic eqution + + 6 0. The middle term is nd coefficient of is. Let us split such tht m + n nd mn 6 + + + 6 0, ( + ) + ( + ) 0 ( + )( + ) 0 Now, + + 6 0 is split into two liner fctors ( + ) nd ( + ). By using zero product rule, we get ( + )( + ) 0 + 0 or + 0, If + 0,, If + 0, the solution set is {, }. Thus, nd re the two roots of the qudrtic eqution + + 6 0. This cn lso e digrmmticlly represented using lgeric tiles s follows. + +6 + + On rerrnging ll the tiles we get the figure Oserve tht the length + nd redth + nd totl re is + + 6. The re of rectngle, A l + + 6 ( + )( + ) From this, we cn conclude tht If the qudrtic polynomil + + c represents the re of squre or rectngle, then the length nd redth represent the two fctors of it. Study the emples on solving qudrtic eqution y fctoristion method. ( + ) ( + )
Qudrtic Equtions 0 ILLUSTRATIVE EXAMPLES Emple : Solve + 0 Sol. Given + 0 + 0 (Resolving the epression) ( ) ( ) 0 (Tking common fctors) ( )( ) 0 (Tking common fctors) 0 or 0 (Equting ech fctor to zero) + or nd + re the roots of + 0 Emple : Solve 6 0 Sol. Given: 6 0 6 + 0, ( + ) ( + ) 0, ( + )( ) 0 + 0 or 0, or or nd re the roots of 6 0 Emple : Find the roots of 6 + 0. Sol. Given: 6 0 6 6 0, 0 0 0 or 0 or, or So, the root is repeted twice, one for ech repeted fctor. The two equl roots of Emple : Solve 0 6 0 re,. Sol. Given: 0 Squring on oth sides, we get 0 ( ) 0 9 + 6, 6 + 0 + 9 0 6 + 0 0, 8( ) + ( ) 0 ( )(8 + ) 0, 0 or 8 + 0 or 8, or 8 nd 8 re the roots of 0.
0 UNIT-9 Steps of finding roots of qudrtic eqution y fctoristion method.. Write the given eqution in stndrd form of qudrtic eqution, + + c 0.. Resolve the qudrtic epression (LHS) y splitting its middle term.. Tke the common fctor nd otin the two liner fctors.. Equte ech fctor to zero.. Simplify ech liner eqution nd find the vlue of unknown. EXERCISE 9. Solve the qudrtic equtions y fctoristion method:. + + 0 0. 0 0. 6 p p. + 0. m 6(m + ) 6. 00 0 + 0 7 7. 7 0 8. + k + k 0 9. m 6 m 0... y 6y +. 0.t 0.t 0.0. + + 6 0.. 6. 7. ( + ) + 0 8. ( + ) ( + ) 9. ( ) + 0. ( ) Completing the squre method Let us consider the qudrtic eqution. + + 0 Here, the middle term cnnot e split into terms such tht m + n nd mn. This mens we cnnot resolve the eqution s product of two fctors nd therefore, it cnnot e solved y fctoristion method. This is the limittion of fctoristion method of solving qudrtic equtions. This method cn e used only when it is possile to split the middle term nd fctorise the given qudrtic polynomil. Then, how to solve qudrtic equtions where the qudrtic polynomil cnnot e fctorised? We cn lso digrmmticlly represent the given qudrtic eqution + + 0 to understnd why it cnnot e resolved into fctors. Oserve the following figure. We see tht the figure representing + + is not complete squre or rectngle. It mens tht, if the figure is complete squre or rectngle then we cn solve it y fctoriztion method. + + Such in complete squres or rectngles cn e converted into complete squre or rectngles y dding some quntities to it. This method of dding quntities to mke it perfect squre/rectngle is clled completing the squre method.
Qudrtic Equtions 0 B. Solving qudrtic eqution y completing the squre method. Consider the eqution + + discussed ove. We see from the digrm tht ( + + ) is or less to form complete squre or rectngle. In other words, {( + + ) + } will form rectngle nd it cn e fctorised. i.e., + + + 0 + + 6 0 ( + )( + ) 0 or Consider nother emple of solving + + 0 The middle term is nd it cnnot e split such tht the sum of the terms is nd the product is +. Hence, + + do not represent complete squre. Wht is the minimum quntity to e dded to given eqution or sutrcted from it, to mke it complete squre? Recll the identity tht represents complete squre. we know, + + ( + )( + ) Thi s in dictes tht ny qudrtic polynomil cn e resolved into liner fctors, if it is in the form ( + + ). In, + + + + ( + ) the middle term is, where is the constnt is squre root of first term. ( + ) is squre root of lst term. In the given eqution + + 0, the middle term is. Compre with. [ ] Oserve tht, which is hlf of the coefficient of. By squring, we get. The LHS of the qudrtic eqution + cn e converted to complete squre y dding to the LHS. If is dded to LHS, the vlue of the eqution chnges. In order to mintin the originl vlue we hve to dd to LHS nd RHS or we cn dd nd sutrct to the LHS. Agin, tke the eqution + + 0 + + + +, + + ( + )
06 UNIT-9 So, solving + + 0 is equivlent to solving ( + ) + 0 or ( + ). This mens, we cn convert qudrtic eqution to the form ( + ) 0 nd then find its roots. In the figure given elow, oserve how ( + ) is eing converted to ( + ) or ( + ). + + + + + ( +) + + + ( + ) + + Now, let us solve + + 0. ( + ) Find, hlf of the coefficient of. By squring it, we get By dding nd sutrcting, 0 ( + ) + 0 ( + ) Tking squre root on oth the sides, We get, + + If, + +, + or +, + nd re the roots of + + 0 So, we hve solved the qudrtic eqution + + 0 y the process of completing the squre. This known s the method of completing the squre. Consider the squre of the inomil Note tht the lst term is the squre of hlf the coefficient of.
Qudrtic Equtions 07 Hence, ( + ) lcks only the term of eing the squre of. Thus, if the squre of hlf the coefficient of is dded to the epression of the form +, the result is the squre of inomil. Such n ddition is known s completing the squre. ILLUSTRATIVE EXAMPLES Emple : Solve the qudrtic eqution + 6 7 0 y completing the squre. Sol. Given: + 6 7 0 6 + 6 + 9 9 7 0 6 ( ) + + + 9 +6 6 ( + ) + ( + ) 6 ( + )( + ) 6 9 ( + ) () Tking squre root on oth the sides, + (Hlf of coefficient of 6 ) If, + + 7 nd 7 re the roots of + 6 7 0 Emple : Solve + 0 y completing the squre method. Sol. Given + 0 Here, the coefficient of is nd it is not perfect squre. Such qudrtic equtions re solved in two wys. Let us do oth of them. (i) Multiply the eqution through out y. ( + 0) 9 + 6 0 Now, hlf of the coefficient of is. nd So, 9 + 6 0 () + 6 6 (Tking squre root on oth the sides) + or
08 UNIT-9 + 6 (ii) Dividing the eqution through out y. ( + 0) 0 Now, let us proceed s erlier. So, 0 6 6 6 6 6 6 6 6 6 By tking squre root on oth the sides, 6 6 + 6 6, 6 6 6 or 6 6 6, 6 6 6 nd re the roots of the qudrtic eqution + 0. Steps of finding the roots of qudrtic eqution y completing the squre method Step : Write the eqution in stndrd form. Step : If the coefficient of is, go to step If not, multiply or divide oth the sides of the eqution y the coefficient of. Step : Find hlf the coefficient of nd squre it. Add this numer to oth the sides of the eqution or dd nd sutrct on LHS of the eqution. Step : Solve the eqution, using the squre root property. If p, then + p or p where, p is non - negtive numer.
Qudrtic Equtions 09 EXERCISE 9. Solve the following qudrtic equtions y completing the squre. (i) 0 + 9 0 (ii) + 0 (iii) + 0 (iv) +6 9 0 (v) + 0 (vi) t + t 7 7 (vii) ( ) ( + 7) (viii) + (i) + 0 () + ( ) 0 We know tht, in mthemtics clcultions nd solving prolems re mde esier y using formule. In the sme wy, qudrtic equtions cn e esily solved using formul. The qudrtic formul, which is very useful for finding its roots cn e derived using the method of completing the squre. Let us derive the qudrtic formul nd lern how to use it for finding roots of the qudrtic equtions. c) Solution of qudrtic eqution y formul method Consider the qudrtic eqution + + c 0, 0. Divide the eqution y (i.e., coefficient of ) Find hlf the coefficient of nd squre it c 0 Trnspose the constnt c to RHS c Add to oth sides of the eqution c Fctorise LHS nd simplify RHS c c Tke squre root on oth sides of the eqution c c The roots of the qudrtic eqution + + c 0 re nd is known s qudrtic formul.
0 UNIT-9 In the ove derivtion, we hve eliminted the coefficient of, which is not perfect squre y dividing the eqution y. However, it cn lso e solved y multiplying the eqution y. This method is lso clled Sridhrchry's method. Sridhrchry (0 A.D) is credited with deriving the formul for solving qudrtic equtions y the method of completing the squre. Study the derivtion of qudrtic formul s evolved y the ncient Indin mthemticin, Sridhrchry. Consider the generl form of qudrtic eqution. + + c 0, where 0. Multiply oth the sides y {( + + c) 0}, + + c 0, Add to oth the sides, we get () + + c, + c + + c, ( + ) c Tking squre not on oth side + c,,, nd ILLUSTRATIVE EXAMPLES Emple : Solve 7 + 0 y formul method. Sol. Given 7 + 0 This is in the form + + c 0 where,, 7 nd c Qudrtic formul is, By sustituting the vlues, we get ( 7) ( 7) 7 9 8 7 7 If 7 8 7 6 nd re the roots of 7 + 0. Emple : Solve m + m Sol. Given m + m Rewrite the eqution in stndrd form i.e., m m 0 This is of the form + + c 0 where,,, c (Qudrtic formul) ( ) ( ) ()( )
Qudrtic Equtions 8 nd re the roots of the qudrtic eqution m + m. Emple : Solve Sol. Given Simplify the eqution. ( ) ( ) ( )( ) ( )( ) ( )( ), ( + )( + ) ( + ), + 6 + 8 + 8 0. It is qudrtic eqution, Here,,, c 8 ( ) ( ) ()( 8) () 8 nd nd re the roots of the given qudrtic eqution. Emple : Solve + ( + )( + ) 0 y using qudrtic formul. Sol. Given + ( + )( + ) 0, Here,,, c ( + )( + ) ( )( ) () ( ) 8 9 ( ) ( ) nd + ( + ) nd The roots of + ( + )( + ) 0 re ( + ) nd ( + )
UNIT-9 Steps for solving qudrtic eqution using the qudrtic formul Step : Write the eqution in stndrd form, + + c 0 Step : Compre the eqution with stndrd form nd identify the vlues of,, c. Step : Write the qudrtic formul Step : Sustitute the vlues of, nd c in the formul. Step : Simplify nd get the two roots. EXERCISE 9. Solve the following qudrtic equtions y using the formul method.. + 0. + 0. 7 + 0. y + 6y. m m + 0 6. 8r r + 7. p p 8. ( + )( ) + 0 9. + ( ) 0 0. 9. ( + ) ( + ). 6 + ( ) 0. 0. 8 So fr we hve lernt to find the roots of given qudrtic equtions y different methods. We see tht the roots re ll rel numers. Wht is the nture of these roots? Wht determines the nture of the roots? Is it possile to determine the nture of roots of given qudrtic eqution, without ctully finding them? Discuss in clss nd try to nswer these questions. Now, let us lern out this. Nture of the roots of qudrtic equtions Study the following emples. Consider the eqution + 0 This is in the form of + + c 0,,, c, ( ) ( ), 0 0 or 0 or roots re equl
Qudrtic Equtions. Consider t he equt ion 0 This is in the form + + c 0, where,, c ( ) ( ) ( ) or, 6 or or - roots re distinct. Consider the eqution + 0 This is in the form + + c 0, where,, c ( ) ( ) ()() 8 or roots re imginry From the ove emples, it is evident tht the roots of qudrtic eqution cn e rel nd equl, rel nd distinct or imginry. Also, oserve tht the vlue of c determines the nture of the roots. We sy the nture of roots depends on the vlues of c. The vlue of the epression c discrimintes the nture of the roots of + + c 0 nd so it is clled the discriminnt of the qudrtic eqution. It is denoted y the symol nd red s 'delt'. In generl, the roots of the qudrtic eqution + + c 0 re If, c 0 then, the eqution hs two equl roots which re rel. Thus, If, c > 0 then, the eqution hs two distinct roots which re rel. Thus,, If, c < 0 then, the eqution hs no rel roots. Since, ( c ) cnnot e found nd we sy it is imginry.
UNIT-9 The ove results re presented in the tle given elow. Discriminnt Nture of roots 0 Rel nd equl > 0 Rel nd distinct < 0 No rel roots (imginry roots) ILLUSTRATIVE EXAMPLES Emple : Determine the nture of the roots of the eqution 0 Sol. This is in form of + + c 0. The co-efficient re,, c c, ( ) () ( ) + 8, > 0. Therefore, roots re rel nd distinct Emple : Determine the nture of the roots of the eqution + 0 Sol. Consider the eqution + 0 This is in the form of + + c 0. The co-efficient re,, c c, ( ) () (), 6 6 0. Therefore, roots re rel nd equl Emple : For wht positive vlues of 'm' roots of the eqution + m + 0 re (i) equl (ii) distinct Sol. Consider the eqution + m + 0 This is in the form + + c 0. The co-efficients re, m, c c, m ()(), m 6 (i) If roots re equl, then 0 m 6 0, m 6, m 6 m (ii) If roots re distinct, then 0 m 6 0, m 6, m 6 m EXERCISE 9.6 A. Discuss the nture of roots of the following equtions (i) y - 7y + 0 (ii) - + 0 (iii) n + n - 0 (iv) + + 0 (v)) + - 0 (vi) d - d + 0 B. For wht positive vlues of m roots of following equtions re ) equl ) distinct ) imginry i) -m+0 ii) -m+90 iii) r -(m+) r+0 iv) mk -k+0 C. Find the vlue of P for which the qudrtic equtions hve equl roots. i) -p+9 0 ii) ++p 0 iii) pk -k+9 0 iv) y -py+ 0 v) (p+) n + (p+)n+(p+8)0 vi) (p+)c +(p+)c+p0
Qudrtic Equtions Reltionship etween the roots nd co-efficients of the terms of the qudrtic eqution. If m nd n re the roots of the qudrtic eqution ++c0 then m nd n m+n + m + n mn mn c mn c If m nd n re the roots of the qudrtic eqution + + c 0 Sum of the roots (m+n) - Product of roots mn +c ILLUSTRATIVE EXAMPLES Emple : Find the sum nd product of the roots of eqution + + 0 Sol. This is in the form + + c 0, where,, c Let the roots e m nd n i) Sum of the roots m+n m n ii) Product of the roots mn c mn Emple : Find the sum nd product of the roots of eqution + 0 Sol. This is in the form + + c 0, where, 0, c Let the roots e p nd q i) Sum of the roots p+q 0 p+ q 0 ii) Product of the roots pq c pq
6 UNIT-9 Emple : Find the sum nd product of the roots of eqution (p + q) + pq 0. Sol. The coefficients re, (p + q), c pq i) Sum of the roots m + n m + n ( p q) m n ( p q) c ii) Product of the roots mn pq mn pq. EXERCISE 9.7 Find the sum nd product of the roots of the qudrtic eqution:. + 8 0. 0 0. 8m m. 6k 0. pr r 6. + () + ( + ) 0 Frming qudrtic eqution : We know how to find the roots, of given qudrtic equtions nd lso their sum nd product. Is it possile to frme qudrtic eqution if sum nd product of its roots re given? If 'm' nd 'n' re the roots then the stndrd form of the eqution is (sum of the roots) + product of the roots 0, i.e. (m + n) + mn 0. ILLUSTRATIVE EXAMPLES Emple : Form the qudrtic eqution whose roots re nd. Sol. Let 'm' nd 'n' e the roots m, n Sum of the roots m + n + m + n Product of the roots mn () () mn 6 Stndrd form is (m + n) + mn 0 () + (6) 0 6 0 Emple : Form the qudrtic eqution whose roots re nd Sol. Let 'm' nd 'n' e the roots m nd n Sum of the roots m + n m n 6 Product of the roots mn () 9 0 mn (m + n) + mn 0 6 0 Emple : If 'm' nd 'n' re the roots of eqution + 0. Find the vlue of (i)m n + mn (ii) m n. Sol. Consider the eqution + 0 This is in the form + + c 0, where,, c
Qudrtic Equtions 7 (i) Sum of the roots m + n ( ) m n (ii) Product of the roots mn c mn Therefore, (i) m n + mn mn (m + n) (ii) m n n m m n mn mn m n Emple : If 'm' nd 'n' re the roots of eqution + 0 form the eqution whose roots re m nd n. Sol. Consider the eqution + 0. Here,, c (i) Sum of the roots m + n ( ) m n (ii) Product of the roots mn c mn If the roots re m nd n Sum of the roots m + n (m + n) mn () () 9 8 m + n Product of the roots m n (mn) m n 6 (m + n ) + m n 0 () + (6) 0 6 0 Emple : If one root of the eqution 6 + q 0 is twice the other, find the vlue of 'q'. Sol. Consider the eqution 6 + q 0, where, 6, c q (i) Sum of the roots m + n ( 6) m n 6 c q (ii) Product of the roots mn mn q If one root is m then twice the root is m. m m nd n m m + n 6 m + m 6, m 6 m 6 We know tht q mn q m(m) m () 8 q 8 Emple 6: Find the vlue of k so tht the eqution + (k + ) 0 hs one root equl to zero. Sol. Consider the eqution + (k + ) 0. Here,,, c k + Product of the roots mn c mn k mn k + Since 'm' nd 'n' re the roots, nd one root is zero then m m nd n 0, mn k + m(0) k + 0 k + k.
8 UNIT-9 A. Form t he equt ion whose root s re EXERCISE 9.8 i), ii) 6, iii), v) ( ), ( ) (vi) ( ), ( ) iv), B.. If 'm' nd 'n' re the roots of the eqution 6 + 0 find the vlue of (i) (m + n) mn (ii) m n (iii) m n n m (iv) n m. If '' nd '' re the roots of the eqution m 6m +, find the vlue of (i) (ii) ( + )( + ). If 'p' nd 'q' re the roots of the eqution + 0. Find the vlue of (i) (p + q) + pq (ii) p + q. Form qudrtic eqution whose roots re p q nd q p. Find the vlue of 'k' so tht the eqution + + (k + ) 0 hs one root equl to zero. 6. Find the vlue of 'q' so tht the eqution q + q 0 hs one root which is twice the other. 7. Find the vlue of 'p' so tht the eqution 8p + 9 0 hs roots whose difference is. 8. If one root of the eqution + p + q 0 is times the other prove tht p 6q. D. Solution of qudrtic eqution y grphicl method. You re fmilir with drwing grphs for liner equtions nd lso solving simultneous liner equtions grphiclly. Now let us drw grphs for qudrtic epressions nd lern to solve gudrtic equtions grphiclly. Y Emple : Consider the qudrtic eqution 0. We need co-ordintes of points to plot them nd drw the grph. Let y Prepre the tle for vlues of '' nd 'y' for the eqution y. y 0 0 Plot the points A(, ), B(, ), C(0, 0), D(, ) nd E(, ). Join the points y smooth curve. The grph of y is curved line. X A(, ) 6 B(, ) D(, ) O C(0, 0) Y E(, ) X
- is : cm unit y - is : cm unit Qudrtic Equtions 9 Oserve the grph of two more qudrtic equtions. Emple : 0 Emple : 0 0 Let y y 8 0 8 X A(-,8) B(,) 9 8 7 6 O Y Let y 0 y ½ 0 ½ I Y Y Thus, we oserve tht the grphs of qudrtic equtions 0, 0 nd 0 re curved lines. This curved line representing qudrtic equtions is clled "Prol". From the grph of these qudrtic equtions we oserve tht: The prol of the grphs re symmetricl with respect to the y-is. The point where the curvture is gretest is clled the verte. The prol tkes turn t this point. Now let us lern more out prol. Consider epressions ( ) nd ( + + )... + + Here is greter thn zero Here is less thn zero Let y Let y + + Prepre tle of vlues. Let us prepre tle of vlues. C(0,0) D(,) E(,8) X X I. P(-,). 0. Q(-,0.) O Y R(0,0) T(,) S(, 0.) X y 0 0 0 Plot the points of ech ordered pir (, y). Drw smooth curve through the plotted points. X A(, ) O B(, ) Y C( ) F(, ) D(, ) E(,0) y 0 6 Plot the points of ech ordered pir (, y). Drw smooth curve through the plotted points. Y D(,6) X 6 C(0, ) E(, ) B(,) A(, ) O Y F(,) X
0 UNIT-9 From the ove two grphs we oserve the following: The prol opens upwrds. As increses y decreses until the minimum vlue of y is reched. The smllest vlue of y is. + + The prol opens downwrds. As increses y lso increses until the miimum point is reched. The highest vlue of y is 6. EXERCISE 9.9 I. Drw the grphs of the following qudrtic equtions: i) y ii) y iii) y + 6 iv) y v) y 8 + 7 vi) y ( + )( ) vii) y + 6 viii) y + Grphicl solutions of qudrtic equtions: In this section let us study the grphicl method of solving qudrtic equtions. ILLUSTRATIVE EXAMPLES Emple : Drw the grph of y nd find its roots. Sol. Let us drw the grph of this eqution nd find the roots grphiclly. Step : Prepre tl e of vl ues to y. y 0 0 0 Step :Plot ll the points on the grph. Step :Drw smooth curve through the plotted points. Step :Mrk the intersecting points of the curve with the -is. The coordintes where the prol intersects the -is re the roots of the eqution. The coordintes of the intercepts re B(, 0) nd E(, 0). The roots of the eqution re nd X Y A(, ) B(, 0) O - intercept C(0 ) Y - is : cm unit y - is : cm unit E(,0) X - intercept D(, ) Verifiction y fctoristion method: 0 + 0 ( + ) ( + ) 0 ( + )( ) 0 + 0 or 0 or
Qudrtic Equtions Emple : Solve the qudrtic eqution 0 + 0 grphiclly. Y Sol. Prepre the tle of vlues for the eqution y 0 +. y 6 6 9 0 Mrk the point t which the curve touches the -is. The prol intersects the -is t only one point, E (,0) The roots of the eqution re nd. X 0 0 y 0 + O A(,6) B(,9) C(,) D(,) E(,0) 6 - intercept X F(6,0) Emple : Drw the grph nd find the roots of y +. Sol. Prepre tle of vlues for the eqution y +. y 0 6 6 The prol does not intersect the -is. There is no rel vlue of for + 0. Hence, there re no rel roots. X A(, 6) Y 8 7 6 Y B(, ) D(, ) C(0, ) O E(, 6) X Let us now record the detils of the ove three emples in the tle. Study them: 0 + + It hs points of It hs point of It hs no point of intersection intersection Intersection Y X X X X X X solutions solution No solutions + + c 0 + + c 0 + + c 0 hs rel nd hs rel nd hs no rel unequl roots repeted roots roots c > 0 c 0 c < 0
UNIT-9 Emple : Solve the eqution y ( )( + ) grphiclly. Sol. Step : The given eqution is y ( )( + ). y Simplify the RHS nd ring it to the stndrd form + + c 0 y ( )( + ) + 8 y + 8 0 0 8 9 8 0 Mrk the intersecting points of the curve wi th the -is. The interesecting points re (,0) nd (,0). Hence the roots re ( nd ) X D(, 9) Y 9 C(, 8) 8 E(0, 8) B(, ) A(, 0) 7 6 O Y F(, ) G(, 0) X Emple : Drw the grph of y nd find the vlue of using the grph. Sol. Step : Prepre the tle of vlues for y y 0 0 8 8 0 Step : Plot ll the points on the grph sheet nd drw the prol Step : When, y 0 Drw the stright line y 0 prllel to -is. Step : From the intersecting points of the prol nd the stright line, drw perpendiculrs to the -is. The point on -is t which the perpendiculrs meets re the vlues X y0 0 9 Y D(, 8) 8 C(, 8) 7 6 B(, ) A(, ).. O X of.. Y ±.. Ect solution Approimte solution
Qudrtic Equtions EXERCISE 9.0 I. Drw the grph of the following equtions. i) y ii) y iii) y iv) y + 8 6 v) y vi) y II.. Drw the grph of y nd find the vlue of 7..Drw the grph of y nd find the vlue of 0. Solving prolems sed on qudrtic equtions. We come cross mny situtions in our dily life where we cn solve them y pplying the methods of solving qudrtic equtions. Let us consider some emples. ILLUSTRATIVE EXAMPLES Emple : The product of two consecutive positive odd numers is 9. Find the numers. Sol. Step : Frming the eqution Let one of the odd positive numer e. The other odd positive numer will e ( + ) The product of the numers is ( + ) 9 + 9 0 Step : Solving the eqution + 9 0 + 9 0 ( + )( ) 0 + 0 or 0 or Step : Interpreting nd finding the solution Since the numers must e positive, is not tken. nd + + the two consecutive odd positive numers re nd. Emple : A mn trvels distnce of 96 km y trin nd returns in cr which trvels t speed of km/hr fster thn the trin. If the totl journey tkes hours, find the verge speed of the trin nd the cr respectively. Sol. Let the speed of the trin e km/hr. Then the speed of the cr is ( + ) km/hour Time tken for journey y trin 96 hours Time tken for journey y cr 96 hours
UNIT-9 Totl journey period Journey period y trin + Journey period y cr hrs. 96 96 96( ) 96 ( ) 96 + 6 + 96 + 6 6 0 ( 6) ( 6) ()( 6) 6 8 or.6 Since the speed should e positive, the verge speed of the trin is 8 km/hr nd the verge speed of the cr is (8 + ) 9 km/hr Emple : Anirudh ought some ooks for Rs. 60. Hd he ought more ooks for the sme mount ech ook would hve cost him rupee less. Find the numer of ooks ought y Anirudh nd the price of ech ook. Sol. Let the numer of ooks e Totl cost of the ooks ` 60 Cost of ech ook ` 60 If the numer of ooks is ( + ), Then, the cost of ech ook ` Difference in cost is one rupee Cost of ech ook when 60 Cost of ech ook when numer numer of ooks is ( ) of ooks is ( + ) 60 60 60( ) 60 ( ) 60 00 60 00
Qudrtic Equtions + 00 + 00 0 + 0 00 0 ( + 0) ( + 0) 0 ( + 0)( ) 0 + 0 0 or 0 0 or Numer of ooks Numer of ooks cnnot e negtive. Hence 0 is rejected. Cost of ech ook 60 60 ` EXERCISE 9.. Find two consecutive positive odd numers such tht the sum of their squres is equl to 0.. Find the whole numer such tht four times the numer sutrcted from three times the squre of the numer mkes.. The sum of two nturl numers is 8. Determine the numers, if the sum of their reciprocls is 8.. A two digit numer is such tht the product of the digits is. When 6 is dded to this numer the digits interchnge their plces. Determine the numer.. Find three consecutive positive integers such tht the sum of the squre of the first nd the product of other two is. 6. The ges of Kvy nd Krthik re yers nd yers. In how mny yers time will the product of their ges e 0. 7. The ge of mn is twice the squre of the ge of his son. Eight yers hence, the ge of the mn will e yers more thn three times the ge of his son. Find their present ge. 8. The re of rectngle is 6 cm. If the mesure of its se is represented y + nd the mesure of its height y, find the dimensions of the rectngle. 9. The ltitude of tringle is 6cm greter thn its se. If its re is 08 cm. Find its se nd height. 0. In rhomus ABCD, the digonls AC nd BD intersect t E. If AE, BE + 7, nd AB + 8, find the lengths of the digonls AC nd BD.
6 UNIT-9. If twice the re of smller squre is sutrcted from the re of lrger squre, the result is cm. However, if twice the re of the lrger squre is dded to three times the re of the smller squre, the result is 0 cm. Determine the sides of the two squres.. In n isosceles tringle ABC, AB BC nd BD is the ltitude to se AC. If DC, BD nd BC +, find the lengths of ll three sides of the tringle.. A motor ot whose speed is km/hr in still wter goes 0 km downstrem nd comes ck in totl of hours 0 minutes. Determine the speed of the strem.. A deler sells n rticle for ` nd gins s much percent s the cost price of the rticle. Find the cost price of the rticle.. Nndn tkes 6 dys less thn the numer of dys tken y Shoh to complete piece of work. If oth Nndn nd Shoh together cn complete the sme work in dys, in how mny dys will shoh lone complete the work? 6. A prticle is projected from ground level so tht its height ove the ground fter t second is given y (0t t )m. After how mny seconds is it m ove the ground? Cn you eplin riefly why there re two possile nswers?
Qudrtic Equtions 7 Qudrtic Equtions Pure qudrtic equtions + c 0 Adfected qudrtic equtions + + c 0 Reltion etween roots nd coefficients Nture of roots Solving qudrtic equtions Fctoristion method Completing the squre method Formul method Sum of roots m + n - Product of roots mn c Frming Q.E. (m n) mn 0 0 > 0 Discriminnt c Roots re rel nd equl Roots re rel nd distinct Grphicl method < 0 Roots re imginry Drwing prol Solving Q.E. y drwing prol Finding,... y using prol EXERCISE 9. ANSWERS ] (i) (ii) (iii) 0 (iv) 8 7 (v) nd (vi) (vii) 9 (viii) 7 ] (i) r (ii) d (iii) (iv) 8 (v) v 0 (vi) v 0 EXERCISE 9. ] 0, ], ], ], ], 6] 0, 0 7], - 8] k, k 9] 7, 0], ], ], ], ] 0, ], 6], 7], 8], 9], 0] 6,
8 UNIT-9 EXERCISE 9. (i) 9, (ii), - (iii), (vii) 9, 0 (viii), (i), () EXERCISE 9. ] ] - ], ] 7] ] - 9 8], - 9] ] 8, (iv) 8 ± 7 (v) ± 0] 0, 8 ], ], 6] 6 (vi) 6 -± 7 6 ] 6 EXERCISE 9.6 C. (i) 6 (ii) 9 8 EXERCISE 9.7 (iii) (iv) (v) 0 - ], 8 ], ], - ] 0, - 8 EXERCISE 9.8 B. ] (i) (ii) (iii) (iv) 7 (vi), ] p p 6], (+) ] (i) - EXERCISE 9. (ii) 9 ] (i) 6 (ii) ] 6] 7] ] 7, 9 ] ], or, ] 6 ] 8, 9, 0 6] 7], 8] cm, cm 9] cm, 8cm 0] 0cm, cm ] cm, 8cm ] 7cm, 7cm, 6cm ] km/hr ] `0 ] dys 6] or